Task 2
Main Page Task 1 Videos ''Part 2 Solar System: '''The elliptical orbit of Pluto has the greatest eccentricity among all planets of our solar system ( e = 0.248). Pluto’s orbit has the sun at one focus and a major axis of 79.6 AU (astronomical units).' ' ' Consider Pluto’s orbit on a giant coordinate grid where x and y are measured in AU and the sun is at the origin. (As shown in the figure above)' ' A) Use the length of the major axis to determine a. a = 79.6/2 a= 39.8 AU B) Use the eccentricity and �� to determine the focal radius ��. (Round to 2 decimal places.) c = e a c = 0.248(39.8) c = 9.87 AU C) Write an equation of the form (x-c)'''''2/a''2'' + y''2/b2'' = 1 to model Pluto’s orbit. (Find ��''''' first.) First c2 =a2 – b2 (9.87)2 = (39.8)2 – b2 b2 =1486.62 b = 38.56 AU Then Equation: (x-h)2/a2 + (y-k)2/ b2= 1 (x-9.87)2 / (39.8)2+ (y-0)2/ (38.56)2= 1 (x-9.87)2 / 1584.04 + y2/ 1486.62 = 1 D) The perigee is when the Pluto is closest to the sun. The apogee is when it is furthest from the sun. what are the distances from the sun at these points? '' ''perigee= a - c '' = 39.8 - 9.87'' '' = 29.93 AU'' apogee= a + c '' = 39.8+ 9.87'' '' = 49.67 AU'' ''Part 3 'Bridge:' ''' '' ''Mohamed and Salim study a drawing of an ornamental bridge such as the one shown at the right. It shows an elliptical arch that spans a narrow strait of water. The arch they are studying can be modeled by the following formula '' '' x''2/182.25 + y2'' /132.25 =1'' ''1. Mohamed wants to know the dimensions of the arch'' '' a. Which term in the equation defines the horizontal axis?'' - because the equation is horizontal ( number under X is greater than the number under Y ) so the the number under x define the horizontal axis (major axis). '' b. Write a comparative statement to show whether the major axis of the arch  is horizontal or vertical.'' - If the number under X is greater than the number under Y so the equation is horizontal and the major axis is equal to square root of the number under X. And if the number under X is less than the number under Y so the equation is vertical and the major axis is equal to square root of the number under Y. and in this equation the major axis equal to square root of the number under X. '' c. Mohamed says that the width of the bridge is 13.5 feet. Is he correct?  Explain.'' - he is correct because the width of the bridge equal to square root of the number under X. √182.25=13.5 '' d.Write an expression for the height of the bridge.'' - the height of the bridge equal to square root the number under Y. √132.25=11.5 ''2.Salim notes that this bridge is hardly high enough to pass under while standing up in a moderate-size boat. If he were to build a bridge, it would be at least 1.3 times as wide and twice as high.'' '' a.Find the vertices and co-vertices of Salim’s bridge design.'' b= 1.3(13.5) = 17.5 a= 2(11.5) = 23 co-vertex 1 = ( h+b,k) = ( 0+17.5,0) = (17.5,0) co-vertex 2 = ( h-b,k) = ( 0-17.5,0) = (-17.5,0) vertex 1 = (h,k+a) = (0,0+23) = (0,23) vertex 2 = (h,k-a) = (0,0-23) = (0,- 23) '' b.Write an equation for the design of Salim’s bridge using his minimum  dimensions. '' The ellipse will be vertical (x-h)2/b2 + (y-k)2/a2=1 '' (x-0)2/17.52 + (y-0)2/232=1'' '' x2/306.25 + y2/529=1'' Part 5 In the 'LORAN '('LO'ng 'RA'nge 'N'avigation) radio navigation system, two radio stations located at A and B transmit simultaneous signals to a ship located at P. The onboard computer converts the time difference in receiving these signals into a distance difference , and this, according to the definition of a hyperbola, locates the ship on one branch of a hyperbola (see the figure). Suppose that station B is located 400 mi due east of station A on a coastline. A ship received the signal from B 1200 microseconds (ms) before it received the signal from A. (a) Assuming that radio signals travel at a speed of 980 ft/ms, find an equation of the hyperbola on which the ship lies. speed= distance / time 980 = d / 1200 d= 980x1200 d=1176000ft 1mile=5280ft x mile = 1176000 ft x=1176000 / 5280 then x= 2450 / 11 PA - PB = 2a 2a= 2450 / 11 a= 1225 / 11 Center ( 0 , 0 ) so A ( -200 , 0 ) , B ( 200 , 0 ) so c= 200 b''2''= c''2''-a''2'' b''2''= c''2''-a''2'' b''2''=200''2 ''- ( 1225 / 11 )2'' b''2 ''=27598.14 and since a''2 ''=12401.85 (x-h)''2 ''/ a''2 - (y-k)2 ''/ b''2 = 1 x''2 /12401.85 - y''2/27598.14 = 1 (b) If the ship is due north of B, how far of the coastline is the ship? x''2 /12401.85 - y''2/27598.14 = 1 x= 200 (200)2 ''/12401.85 - y''2/27598.14 = 1 3.225'' '' - y''2''/27598.14 = 1 - y''2''/27598.14 = 1- 3.225 - y''2''/27598.14 = -2.225 y''2'' =61414.8y =247.8 or almost y= 248 Mile Main Page '''Task 2